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Geometric Progressions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSGeometric Progressions3 Marks
Question
If a, b, c, d and p are different real numbers such that: $(\text{a}^2+\text{b}^2+\text{c}^2)\text{p}^2-2(\text{ab}+\text{bc}+\text{cd})\text{p}+(\text{b}^2+\text{c}^2+\text{a}^2)\le0,$ then show that a, b, c and d are in G.P.
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Answer

$(\text{a}^2+\text{b}^2+\text{c}^2)\text{p}^2-2(\text{ab}+\text{bc}+\text{cd})\text{p}+(\text{b}^2+\text{c}^2+\text{a}^2)\le0,$ $\Rightarrow(\text{a}^2\text{p}^2+\text{b}^2\text{p}^2+\text{c}^2\text{p}^2)-2(\text{a}\text{b}\text{p}+\text{b}\text{c}\text{p}+\text{c}\text{d}\text{p})+(\text{b}^2+\text{c}^2+\text{d}^2)\le0$ $\Rightarrow(\text{a}^2\text{p}^2-2\text{a}\text{b}\text{p}+\text{b}^2)+(\text{b}^2\text{p}^2-2\text{b}\text{c}\text{p}+\text{c}^2)+(\text{c}^2\text{p}^2-2\text{c}\text{d}\text{p}+\text{d}^2)\le0$ $\Rightarrow(\text{a}\text{p}-\text{b})^2+(\text{b}\text{p}-\text{c})^2+(\text{c}\text{p}-\text{d})^2\le0$ $\Rightarrow(\text{a}\text{p}-\text{b})^2+(\text{b}\text{p}-\text{c})^2+(\text{c}\text{p}-\text{d})^2=0$ $\Rightarrow(\text{a}\text{p}-\text{b})^2=0$ $\Rightarrow\text{p}=\frac{\text{b}}{\text{a}}$ Also, $(\text{bp}-\text{c})^2=0$ $\Rightarrow\text{p}=\frac{\text{c}}{\text{b}}$ Similiarly, $\Rightarrow(\text{cp}-\text{d})^2=0$ $\Rightarrow\text{p}=\frac{\text{d}}{\text{c}}$ $\therefore\frac{\text{b}}{\text{a}}=\frac{\text{c}}{\text{b}}=\frac{\text{d}}{\text{c}}$ Thus, a, b, c and d are in G.P.

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